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3 years ago

Please help quickly as possible thank you :)

Chemistry

1 answer:

WARRIOR [948]3 years ago

4 0

Answer:

moving object transfers some to the stationary object causing it to move a it. remember that momentum is always conserved though - it is the same at the start before the event and after it.

Explanation:

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How many moles of Si are in 78.0 grams of silicon

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The answer is 2.78 moles of Si

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2 years ago

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Write a sentence that descrices how to determine the number of moles of a compound in known mass of a

sattari [20]

Explanation:

<u>Moles is denoted by given mass divided by the molecular mass , </u>

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

For example ,

For a compound X ,

The given mass i.e. w = 20 g

and the molecular mass ,i.e. , m = 10 g / mol

Then the moles can easily be calculated by using the above formula ,

n = w / m

n = 20 g / 10 g/mol = 2 mol

Hence , answer = 2 mol.

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3 years ago

5. In which group of elements do most atoms have completely filled s and p velence sublevels? A) halogens B) noble gases C) alka

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3 years ago

Zinc + Hydrochloric Acid → Zinc Chloride and Hydrogen Gas

Kipish [7]

Explanation:

because in the right side there is 2 hydrogen( H) and 2 chloride (Cl) to balace them we have to replace 2.

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3 years ago

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One of the emission spectral lines for Be3+ has a wavelength of 253.4 nm for an electronic transition that begins in the state w

4vir4ik [10]

Answer:

Explanation:

Relationship between wavenumber and Rydberg constant (R) is as follows:

$wave\ number=R\times Z^2\frac{1}{n_1^2} -\frac{1}{n_1^2}$

Here, Z is atomic number.

R=109677 cm^-1

Wavenumber is related with wavelength as follows:

wavenumber = 1/wavelength

wavelength = 253.4 nm

$wavenumber=\frac{1}{253.4\times 10^{-9}} \\=39463.3\ cm^{-1}$

Z fro Be = 4

$39463.3=109677\times 4^2(\frac{1}{n_1^2} -\frac{1}{5^2})\\39463.3=109677\times 16(\frac{1}{n_1^2} -\frac{1}{5^2})\\n_1=4$

Therefore, the principal quantum number corresponding to the given emission is 4.

6 0

3 years ago

Please help quickly as possible thank you :) ​

Please help quickly as possible thank you :)